School-Safe Puzzle Games

## How many people in the same room share the same birthday?

Back in the 7th grade, our science teacher Mr. Dabney made a bet with us the first day of school. There were about 35 students in the class, and he wagered 35 boxes of donuts there were at least 2 of us that shared the same birthday (he promised he didn’t know this information beforehand).

We took him up on the bet, confident he was making a foolish mistake, but as Mr. Dabney made his way around the room asking our birthdays, a match popped up just as he got to the last couple of students.

So the question is: How many people do you need to have in a room so that there is a greater than 50% chance that 2 of them will share the same birthday?

How many people do you need to have a 97% chance that 2 will share the same birthday?

### 9 Comments to “How many people in the same room share the same birthday?”

1. Joe | Guest

Not entirely sure, but I think it’s 14 an 19, respectively.

2. doug | Guest

this reminds me of something interesting that happened to me last month…. i work in a restaurant, and therefore often check IDs when young people order alcohol. i checked the ID of a young lady and said ‘hm, you have the same birthday as my mom’–then i checked the guys ID and was suprised to see he shared the same birthday as my sister! i figured the odds of one birthday being shared were 1 in 365, (365.25 if you want to be really accurate) and the odds of 2 birthdays matching would be 365.25X365.25 or 1 in 133,407. kinda random. i guess this would be the same odds as 3 strangers sharing a single birthday….

as for the problem at hand, im not sure. im not quite sure how to set up such an equation. it seems to me that to ensure that there is a 50-50 chance of 2 people sharing a birthday, at least half the days would have to be accounted for, but if you had 182 people, it seems better than a long shot that 2 would match….and jr. high teachers wouldnt be placing such wagers. ill have to think about this. way to make me lose sleep dr. kaplan!!

great question.

3. max | Guest

X=average number of days in a month

you would have to have at least x amount of people in a room to have atleast a 50% chance. so the equation would be 2 out of X. or 1 out of X divided by two. to have atleast a 97% chance, you would have to have twice as many kids as X. so this equation would be 2 out of two times X, or 1 out of X.

^thats just what i think.

4. Scott | Guest

Well its a matter of statistics. since the first birthday doesnt matter, we can use 1/1 as the fraction for that, and 1/365 since it has to be on the same day as the first for the second.

so 365/2 is the number of people in the same room youd need to have to have a 50% chance.

and .97 times 365 for a 97% chance.

5. max | Guest

^^sorry dad, but the 35 kids in that room is much less than 365/2

6. Elena | Guest

The chance that someone is born on any day in a year is: 1/365
Since the birthday event is independent of any other b-day event…the chance that any 2 people have the chance of being born the same day is:
(1/365)*(1/365)….and yet again one has to consider the chance that this happens on any one day choosing 1 day out of 365 = 365 possible dates and in this class there are 35 potential candidates meaning that there are 2 out of 35 possible combinations….

the solution to the possibility of 2 people sharing the same B-day should thus be:365* 1190*(1/365)*(1/365)

where

365= results from 365 chose 1 day
1190= results from 35 chose 2 candidates

bigger 50% –> 15 candidates
bigger 97%–> 20 candidates

7. Miraculous Paper Cutting Puzzle: Pass a Person Through a Card - Smartkit Brain Enhancement News | Guest

[…] If I were a teacher, I’d offer my class a morning of free donuts if anyone in the class  figured this puzzle out : ) […]

8. 8DX | Profile

The chance of at least 2 people in a given group having the same birthday is over 50% at 23 people in the group and at 50 people it reaches 97%. For 35 people it will be about 80% so that’s quite a good bet.